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For any continuous map between the displayed Real projective spaces, the induced map on cohomology sends the degree-one generator to $\alpha x+\beta y$. Its $2n$th power is $\left(\genfrac{}{}{0ex}{}{2n}{n}\right){\alpha }^n{\beta }^n{x}^n{y}^n$, since all other terms exceed the truncation exponent of one factor. The coefficient is zero modulo two because $\left(\genfrac{}{}{0ex}{}{2n}{n}\right)=2\left(\genfrac{}{}{0ex}{}{2n-1}{n-1}\right)$. Naturality of the evaluation pairing then makes the top induced map on homology zero. This obstruction does not apply to $n=0$, when both spaces are points.

The chain map identity $d_{D}f=f{d}_C$ gives, for $(x,y)\in M_i$,Thus $M$ is a chain complex; it is a sign convention for the algebraic mapping cone.

Put $E_i=C_{i-1}$ with differential $d_E=d_C$. Inclusion into the second summand and projection onto the first give a short exact sequence of chain complexesIts connecting homomorphism is particularly explicit. If $x\in C_{i-1}$ is a chain cycle, lift it to $(x,0)\in M_i$ and take its boundary. The second component is $(-1{)}^{i}f(x)$, so the long exact sequence in homology containsIf every $H_i(M)$ vanishes, the exact sequence makes every $f_{\ast }$ both injective and surjective. Conversely, if every $f_{\ast }$ is an isomorphism, the map $H_i(D)\to H_i(M)$ has zero image, while $H_i(M)\to H_{i-1}(C)$ has zero image and zero kernel; hence $H_i(M)=0$.

One can also see the converse directly on chain cycles. If $(x,y)\in M_i$ is a chain cycle, then $d_{C}x=0$ and $d_{D}y=-(-1{)}^{i}f(x)$. Injectivity of the induced map on homology $f_{\ast }$ gives $x=d_{C}z$. Subtracting the chain boundary $d_f(z,0)$ leaves $(0,t)$ with $d_{D}t=0$. Surjectivity of the induced map on homology $f_{\ast }$ lets us choose a chain cycle $w\in C_i$ and $v\in D_{i+1}$ such thatThen $(0,t)=d_f(w,v)$. Thus every chain cycle in $M$ is a chain boundary, proving the mapping cone acyclicity criterion:

Let $f:(C,d)\to (C^{\prime },d^{\prime })$ be a chain map, soIf $x\in C_i$ is a cycle, then $d^{\prime }{f}_i(x)=f_{i-1}d(x)=0$, so $f_i(x)$ is a cycle. If $x=d_{i+1}y$ is a boundary, thenis a boundary. Henceis a well-defined induced map on homology.

The maps $f$ and $g$ are chain homotopic if there are homomorphismssuch thatFor a cycle $x$, this giveswhich is a boundary. Thus $f_{\ast }[x]=g_{\ast }[x]$ and $f_{\ast }=g_{\ast }$ on homology.

Now consider the proposed mapping cone $M(f)$. Applying its differential twice givesThe diagonal entries vanish because $C$ and $C^{\prime }$ are chain complexes, and the lower-left entry isby the chain-map identity. Therefore $d_f^2=0$ and $M(f)$ is a chain complex.

There is a short exact sequence of chain complexeswhereIts long exact sequence in homology isTo identify the connecting map, represent a class in $H_{i-1}(C)$ by a cycle $x$ and lift it to $(x,0)\in M(f{)}_i$. ThensoAt the preceding occurrence the degree is $i+1$, giving $(-1{)}^{i+1}{f}_{\ast }$. Hence the sequence is exactly

Finally suppose $f-g=d^{\prime }h+hd$. DefineA direct calculation givesThese are equal precisely because $f-g=d^{\prime }h+hd$. Thus $\Phi$ is a chain map. Replacing the plus sign in its definition by a minus sign gives its inverse, so $M(f)$ and $M(g)$ are isomorphic as chain complexes.